hdu2120 Ice_cream's world I (并查集查找环的个数)

Ice_cream's world I
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1747    Accepted Submission(s): 1017

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition
the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and
B are distinct). Terminate by end of file.

Output

Output the maximum number of ACMers who will be awarded.

One answer one line.

Sample Input

8 10

0 1

1 2

1 3

2 4

3 4

0 5

5 6

6 7

3 6

4 7

Sample Output

3

#include <stdio.h>
int fa[1005];
int find(int x)
{
if(fa[x]!=x) fa[x]=find(fa[x]);
return fa[x];
}
bool comp(int a,int b)
{
a=find(a);
b=find(b);
if(a==b) //如果根相等就是出现了环
return true;
else
{
fa[a]=b;
return false;
}
}
void init(int n)
{
for(int i=0;i<n;i++)
fa[i]=i;
}
int main()
{
int n,k;
while(scanf("%d %d",&n,&k)!=EOF)
{
init(n);
int sum=0;
for(int i=0;i<k;i++)
{
int a,b;
scanf("%d %d",&a,&b);
if( comp(a,b) )
sum++;
}
printf("%d\n",sum);
}
return 0;
}